2
$\begingroup$

While discussing witnesses, in https://arxiv.org/abs/0811.2803 the authors mention (page 16 of the arxiv version, below Eq. (32)) that a necessary condition for a witness $W$ to be optimal is that it touches the set of separable states, meaning there's some separable $\rho_s$ such that $\operatorname{tr}(\rho_s W)=0$. They then mention that this condition is not sufficient for optimality, and a witness satisfying this condition is sometimes called "weakly optimal".

From a geometric perspective, this seems weird to me: I imagine witnesses as hyperplanes in state space separating separable states $\mathcal C$ from the rest, as per the hyperplane separation theorem. So if a witness corresponds to a hyperplane touching $\mathcal C$, it seems intuitive that it should be optimal, meaning the plane can't be "made closer" to $\mathcal C$. On the other hand, it might just be possible that one can move the plane in a direction "orthogonal" to the one touching $\mathcal C$. It's just hard to imagine things in such high dimensions.

What are examples of witnesses that are "weakly optimal" in the above sense?

$\endgroup$
6
  • $\begingroup$ I guess one option would be if the witness touches the set of separable states at the boundary of the space of all states; then one could imagine tilting the hyperplane without loosing states (as the "states" one looses would be unphysical, i.e. non-positive matrices). -- I suspect one should be able to cook up an example based on that. $\endgroup$ Commented Mar 11 at 18:25
  • $\begingroup$ @NorbertSchuch mh, interesting. How do you formalise "tilting the hyperplane" though? I'm not sure how to deal wth variations of a given witness still ensuring its block positivity. I eventually worked out an example that seems to work, see below. It's a witness that's zero on $|01\rangle$ and $|10\rangle$, but not $|22\rangle$, and can be made finer moving along the $|22\rangle$ direction. I'm still a bit confused on how to picture the geometry here though $\endgroup$
    – glS
    Commented Mar 11 at 23:18
  • $\begingroup$ I think geometrically it is clear what tilting a plane means. When you touch the separable states, the only way you can change the plane given by the witness is to keep that point fixed or "roll the plane off" around this point, which I would call "tilting" (i.e. it is not a parallel shift). Then you loose states (or rather points) detected by the witness. But if those points are outside the set of physical states, all is good. (This only works at "pointy" points since otherwise you can't tilt and keep the point fixed.) $\endgroup$ Commented Mar 12 at 1:33
  • 1
    $\begingroup$ The geometrical aspect of the question can be made clear with an analogy in 2d: imagine that the set of all states is the square $[0,1]\times[0,1]$, and the set of separable states is the triangle with vertices $(0,0),(0,1)$, and $(1,0)$. Then an example of a weakly optimal witness is a line passing through the points $(0,1)$ and $(1,1/2)$, and an optimal witness would be a line passing through the points $(0,1)$ and $(1,0)$. $\endgroup$ Commented Mar 13 at 8:59
  • $\begingroup$ @MateusAraújo that picture works, but I wonder whether that kind of thing is really possible for quantum states. State space seems much more symmetric than what you'd need for that example. Though in fairness, analogous reasoning would work even if separabes are say a square inside a circle of states. Algebraically, as per results in linked post, it'd mean some $P\ge0$ such that $\operatorname{tr}(P\rho)=0$ with $\rho$ the "contact (separable) state" and $W-P$ is still a witness. $\endgroup$
    – glS
    Commented Mar 13 at 9:08

2 Answers 2

1
$\begingroup$

Consider the optimal witness $$W = I-2|\psi^-\rangle \langle \psi^-|,$$ and now add some amount of $|01\rangle\langle01|$ to it, i.e., let $$W' = W + \alpha |01\rangle\langle01|,$$ for some $\alpha \in (0,2)$. Now this is clearly still a witness, since we just added a positive operator to it, and it still touches the set of separable states at $|10\rangle$, so it is still weakly optimal, but it is no longer optimal, as $\langle 01 |W' |01\rangle = \alpha > 0$.

Geometrically, we can think of this addition as tilting $W$ away from the set of separable states, while keeping fixed the contact point with $|10\rangle$.

$\endgroup$
0
0
$\begingroup$

The following seems to work: consider in $\mathbb{C}^3\otimes\mathbb{C}^3$ the operator: $$W \equiv \frac{I - 2\mathbb{P}(|\Psi^-\rangle)}{2} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix},$$ where $\{|0\rangle,|1\rangle,|2\rangle\}$ span each of the two qutrit spaces. This $W$ is a witness because it has the form $W=\alpha I-\mathbb{P}(|\Psi^-\rangle)$ with $\mathbb{P}(|\Psi^-\rangle)$ projection onto $|\Psi^-\rangle\equiv\frac{|01\rangle-|10\rangle}{\sqrt2}$, and $\alpha=\frac12$ the largest (squared) Schmidt coefficient of $|\Psi^-\rangle$. Its action is trivial on all computational basis states except for $$W|01\rangle=|10\rangle, \qquad W|10\rangle=|01\rangle.$$

It's also (at least) weakly optimal because $\langle \mathbb{P}_{01},W\rangle=\langle \mathbb{P}_{10},W\rangle=0$, and it's a proper witness because it detects $|\Psi^-\rangle$ as entangled.

However, it's not optimal, because denoting with $P_W$ the set of product states $|e,f\rangle$ such that $\langle \mathbb{P}_e\otimes\mathbb{P}_f,W\rangle=0$, as per the results mentioned here, one can observe that $|2,2\rangle$ isn't in the span of $P_W$. That's because for $|2,2\rangle$ to be spanned by vectors in $P_W$ one would need vectors in there of the form $|e\rangle\otimes|f\rangle$ where both $|e\rangle$ and $|f\rangle$ have nonzero overlap with $|2\rangle$. But then, expanding the expectation value, that would produce terms of the form $|\langle 2,f\rangle\langle 2,e\rangle|^2$, which can't be zero. Thus $|2,2\rangle$ is not spanned by $P_W$.

The projection $P\equiv \mathbb{P}_{2,2}\equiv \mathbb{P}_2\otimes\mathbb{P}_2$ is then a positive semidefinite nonzero operator such that $$W-P = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$ is still a witness. Hence $W$ isn't optimal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.